Two point charges `+8q` and `-2q` are located at `x=0` and `x=L` respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is

Options

A. 8 L
B. 4 L
C. 2 L
D. `L//4`

Answer

To find the location on the x-axis where the net electric field due to the two point charges \( +8q \) and \( -2q \) is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- Charge \( +8q \) located at \( x = 0 \)
- Charge \( -2q \) located at \( x = L \)

### Step 2: Determine the Region for Zero Electric Field
The electric field due to a positive charge is directed away from the charge, while the electric field due to a negative charge is directed towards the charge. Since \( +8q \) is stronger than \( -2q \), the point where the electric field is zero must be located outside the region between the two charges.

### Step 3: Set Up the Electric Field Equations
Let the point where the electric field is zero be at \( x = a \), where \( a > L \). The electric field \( E \) due to each charge can be expressed as follows:

1. Electric field due to \( +8q \):
\[ E_{8q} = \frac{k \cdot 8q}{a^2} \]
(directed to the right)

2. Electric field due to \( -2q \):
\[ E_{-2q} = \frac{k \cdot 2q}{(a - L)^2} \]
(directed to the left)

### Step 4: Set the Magnitudes Equal
For the net electric field to be zero, the magnitudes of the electric fields must be equal:
\[ \frac{k \cdot 8q}{a^2} = \frac{k \cdot 2q}{(a - L)^2} \]

### Step 5: Simplify the Equation
We can cancel \( k \) and \( q \) from both sides:
\[ \frac{8}{a^2} = \frac{2}{(a - L)^2} \]

Cross-multiplying gives:
\[ 8(a - L)^2 = 2a^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 8(a^2 - 2aL + L^2) = 2a^2 \]
This simplifies to:
\[ 8a^2 - 16aL + 8L^2 = 2a^2 \]
Rearranging gives:
\[ 6a^2 - 16aL + 8L^2 = 0 \]

### Step 7: Solve the Quadratic Equation
We can use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( A = 6 \), \( B = -16L \), and \( C = 8L^2 \).

Calculating the discriminant:
\[ D = (-16L)^2 - 4 \cdot 6 \cdot 8L^2 = 256L^2 - 192L^2 = 64L^2 \]

Now substituting into the quadratic formula:
\[ a = \frac{16L \pm 8L}{12} \]
This gives us two solutions:
1. \( a = \frac{24L}{12} = 2L \)
2. \( a = \frac{8L}{12} = \frac{2L}{3} \)

### Step 8: Determine Valid Solution
Since \( a \) must be greater than \( L \), we discard \( \frac{2L}{3} \) as it is less than \( L \). Thus, the valid solution is:
\[ a = 2L \]

### Conclusion
The location on the x-axis at which the net electric field due to the two point charges is zero is at \( x = 2L \).

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